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# AQA Physics PHYA5 - Thursday 18th June 2015 [Exam Discussion Thread] watch

1. Could anyone help me on 2cii please. Not too sure how to go about it. http://filestore.aqa.org.uk/subjects...1-QP-JUN12.PDF

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2. (Original post by Jed-Singh)
Can someone help me with 3bi on june 2012, no idea where that method comes from
It's the ratio of the area of the detector compared with the area of where the radiation will go. ie. the surface area of a sphere (4pir^2)

Fancy explaining part Bii to me?
3. (Original post by JJBinn)
It's the ratio of the area of the detector compared with the area of where the radiation will go. ie. the surface area of a sphere (4pir^2)

Fancy explaining part Bii to me?
sorry I meant 3bii too, I don't understand how that equals activity lol
4. could anyone explain me how to do 4c june 11 please? I don't understand the mark scheme.
5. How's everyone going about today then?
6. (Original post by JJBinn)
Thie iron finishes at 84 but the Molten lead does not finish at 84 though if it's change in temperature is 327-84? That would be 243? So confused haha.

Just realised how much of an idiot I am, that's the difference between 84 and 327 ahaha what a bad mistake
No worries!

(Original post by 000alex)
Don't refracting telescopes too?
No. They suffer from chromatic aberration.

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7. (Original post by Jimmy20002012)
Could anyone help me on 2cii please. Not too sure how to go about it. http://filestore.aqa.org.uk/subjects...1-QP-JUN12.PDF

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Dude I replied to you six pages back :L

http://www.thestudentroom.co.uk/show...7#post57091547

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8. (Original post by Dante991)
How's everyone going about today then?
I've gone through section A of the CGP revision guide, and now doing all the section A papers again. Got two more section As to do.

Then afternoon I'll go through section B in the text book and do all the astro questions
9. (Original post by Dante991)
How's everyone going about today then?
Ngl it kinda looks like everyone's revision today is frantically asking questions in here. 😂

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10. (Original post by JJBinn)
It's the ratio of the area of the detector compared with the area of where the radiation will go. ie. the surface area of a sphere (4pir^2)

Fancy explaining part Bii to me?
If I remember correctly, one in 400 of the photons incident on the detector are actually picked up.

So the activity ACTUALLY incident on the detector is 400 x 0.62.

This means that, using the ratio of number of photons incident on the detector over the number produced by the source, divide the activity actually incident on the detector by the ratio you've just worked out:

(0.62 x 400) divided by 4x10^-3 or whatever the ratio was. Hope that helps

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11. (Original post by Jed-Singh)
sorry I meant 3bii too, I don't understand how that equals activity lol
I've just had a go at explaining it

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12. (Original post by CD223)
No worries!

No. They suffer from chromatic aberration.

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In my notes it says refracting telescopes suffer from chromatic and spherical whereas reflecting only suffer from spherical :\
13. (Original post by 000alex)
In my notes it says refracting telescopes suffer from chromatic and spherical whereas reflecting only suffer from spherical :\
Oh does it? I could be wrong. I've just always been taught the opposite. What notes are they from?

EDIT: I guess you could have "spherical" aberration if the curvature of the lens is not designed properly? Is that what your notes say?

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14. (Original post by Fvthoms)
You have more faith in my abilities than I do . Thank you though, and good luck for tomorrow and FP2!
Thank you!

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15. (Original post by CD223)
Oh does it? I could be wrong. I've just always been taught the opposite. What notes are they from?

EDIT: I guess you could have "spherical" aberration if the curvature of the lens is not designed properly? Is that what your notes say?

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Notes are vague, they only say that it is easier to correct in mirrors.
Googling tells me that only lenses of perfect form don't have problems with spherical aberration.
16. (Original post by 000alex)
Notes are vague, they only say that it is easier to correct in mirrors.
Googling tells me that only lenses of perfect form don't have problems with spherical aberration.
Yeah I wasn't sure whether it meant a "perfect" lens or not. I guess they do suffer from spherical if they aren't cut properly but I'm not sure - I've never seen it on papers.

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17. (Original post by CD223)
If I remember correctly, one in 400 of the photons incident on the detector are actually picked up.

So the activity ACTUALLY incident on the detector is 400 x 0.62.

This means that, using the ratio of number of photons incident on the detector over the number produced by the source, divide the activity actually incident on the detector by the ratio you've just worked out:

(0.62 x 400) divided by 4x10^-3 or whatever the ratio was. Hope that helps

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Yeah that makes sense now, thanks
18. (Original post by CD223)
Does this help?

Attachment 429987

There are initially 3 x 10^22 atoms of U.

The half life of U is 4.5 x 10^9 years so there must be 1.5 x 10^22 atoms of U and 1.5 x 10^22 atoms of Pb after one half life. This is where they intersect on the graph.

After another half life, the U curve halves again. Conversely, the Pb curve doubles. They are the mirror of one another.

EDIT: lol, just seen its part two. Ignore me. I'll get on the case.

...

Right, you know that when there are twice as many U atoms as Pb atoms, the ratio is 2:1. This means that the number of U atoms is two thirds the total number.

So the number of U atoms is:

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Sorry must have not seen it! Not sure where the 2/3 comes from though from the ratio of 1:2?

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19. (Original post by JJBinn)
Yeah that makes sense now, thanks
No worries

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20. (Original post by Dante991)
How's everyone going about today then?
I've got two full papers left to do (hopefully I'll get through them both but at least one!), and my notes to read through. Going to practice writing out derivation of gas pressure as well to see if I can actually remember it.

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